Solve the following logarithmic equation.

log Subscript 3 Baseline left parenthesis x plus 11 right parenthesis plus log Subscript 3 Baseline left parenthesis x plus 3 right parenthesis equals 2log3(x+11)+log3(x+3)=2

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Part 1

Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

A.

The solution set is

StartSet nothing EndSetenter your response here.

​(Simplify your answer. Type an exact answer. Use a comma to separate answers as needed.

B.

There is no solution.

We are given the logarithmic equation:

\[
\log_3 (x + 11) + \log_3 (x + 3) = 2
\]

### Step 1: Apply the logarithmic property
We can use the property of logarithms that states:

\[
\log_b (a) + \log_b (c) = \log_b (a \cdot c)
\]

Using this property:

\[
\log_3 ((x + 11)(x + 3)) = 2
\]

### Step 2: Convert the logarithmic equation to an exponential equation
The equation now becomes:

\[
\log_3 ((x + 11)(x + 3)) = 2
\]

We can rewrite this in exponential form. Since \( \log_b (a) = c \) implies \( b^c = a \), we get:

\[
(x + 11)(x + 3) = 3^2
\]

\[
(x + 11)(x + 3) = 9
\]

### Step 3: Expand the left-hand side
Expand the left-hand side of the equation:

\[
(x + 11)(x + 3) = x^2 + 3x + 11x + 33 = x^2 + 14x + 33
\]

So the equation becomes:

\[
x^2 + 14x + 33 = 9
\]

### Step 4: Simplify the equation
Subtract 9 from both sides:

\[
x^2 + 14x + 33 - 9 = 0
\]

\[
x^2 + 14x + 24 = 0
\]

### Step 5: Solve the quadratic equation
We solve the quadratic equation:

\[
x^2 + 14x + 24 = 0
\]

We can use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \(x^2 + 14x + 24 = 0\), the coefficients are \(a = 1\), \(b = 14\), and \(c = 24\). Plugging these values into the quadratic formula:

\[
x = \frac{-14 \pm \sqrt{14^2 - 4(1)(24)}}{2(1)}
\]

\[
x = \frac{-14 \pm \sqrt{196 - 96}}{2}
\]

\[
x = \frac{-14 \pm \sqrt{100}}{2}
\]

\[
x = \frac{-14 \pm 10}{2}
\]

Now, calculate the two possible values for \(x\):

\[
x = \frac{-14 + 10}{2} = \frac{-4}{2} = -2
\]
\[
x = \frac{-14 - 10}{2} = \frac{-24}{2} = -12
\]

### Step 6: Check for extraneous solutions
Since the logarithm is undefined for negative arguments, we must check whether these values of \(x\) make the expressions inside the logarithms positive.

- For \(x = -2\), \(x + 11 = 9\) and \(x + 3 = 1\), which are both positive, so this solution is valid.
- For \(x = -12\), \(x + 11 = -1\), which is negative, so this solution is **not valid**.

### Final Answer
The only valid solution is \(x = -2\).

The solution set is:

\[
\{-2\}
\]

The correct choice is **A**.

---

Would you like more details on any part of the process?

Here are 5 questions related to this topic:
1. How do you solve logarithmic equations that involve products?
2. Can all logarithmic equations be solved by converting to exponential form?
3. How do you determine whether a logarithmic solution is extraneous?
4. What happens if the argument of a logarithmic function is negative?
5. How does the quadratic formula help in solving logarithmic equations?

**Tip**: Always check for extraneous solutions when solving logarithmic equations, as the logarithm of a negative number is undefined.

Solve the following logarithmic equation.
log Subscript 3 Baseline left parenthesis x plus 11 right parenthesis plus log Subscript 3 Baseline left parenthesis x plus 3 right parenthesis equals 2

A. The solution set is { }.
B. There is no solution.

Learn how to solve the logarithmic equation log3(x + 11) + log3(x + 3) = 2 with a step-by-step solution. This problem covers logarithmic properties, exponential equations, and solving quadratic equations using the quadratic formula. Suitable for students in Grades 9-12.

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